Hi I have just calculated my first attempt at a design of a log spiral antenna and have included all my designs and calculations below. Please could i have feedback on them from any experienced members who have designed these before thanks. Also how do I calculate my diameter ?

Equation of a log spiral is
r1 = RoeaØ [Equation 1]
Where Ro is the initial inner radius and is a constant which controls the initial radius of the spiral. a is known as the expansion coefficient and determines the rate of spiral. According to [1] a is usually between 0.2 and 1.2 and according to [2] a commonly used value is 0.22 and is determined using the pitch or the winding angle ѱ which is calculated as
Tan ѱ = 1/a
So if a is chosen to be 0.22 then our winding angle is
Tan ѱ = 1/0.22
Ѱ = Tan-1 1/0.22
Ѱ = 77.6º or 1.35 radians

The outer edge of the spiral can now be obtained by rotating equation 1 by δ= π/2 or 1.57 radians to make it self- complementary. This allows us to define a second edge but like the one in equation 1 except that it is rotated through an angle of π/2 or 1.57 radians
r2 = Roea (Ø-δ) [Equation 2]
To make the other side of the antenna symmetric we can define the other two edges from the equations below
r3 = Roea (Ø-π) [Equation 3]
r4 = Roea (Ø-π-δ) [Equation 4]

According to [2] the constant in equation 1 Ro which is the inner radius can be calculated from the equation below. The highest frequency in the antennas operating band is said to occur when the innermost radius (where the spiral starts after the feed structure) is equal to λ/4 which should be 0.3m/4 = 0.075m

[Equation 3]
1x109 = 3x108 / 4Ro
4x109 Ro = 3x108
Ro = 3x108/4X109
Ro = 0.075m
Or Ro = 75mm
The total length of the spiral or outer radius can now be calculated from the equation below. The lowest operating frequency is said to occur when the wavelength is equal to the circumference of the spiral so from below 2πRspiral should be equal to the lowest wavelength 1m which would be 0.159x2π which does equal to approximately 1m
[Equation 4]
300x106 = 3x108/2πRSpiral
300x106 x2πRspiral = 3x108
18.85x108Rspiral = 3x108
Rspiral = 3x108/18.858
Rspiral = 0.159m
Or Rspiral = 159mm

If we now insert our value for a, Ro and Ø in to the equations then
r1= 0.075e0.22*1.35
r =0.075e0.3
r1 = 0.101m
Or 101mm

r2 = 0.075e (1.35-1.57)
r2 = 0.075e-0.22
r2 = 0.060m
Or 60mm
According to [2] it is possible to define a ratio for r2 to r1 k and this value of k should fall in the region of 1 to e-πa so we expect k to be in the region of 1 to 0.5 k shown below meets this requirement and can be calculated as shown below
k = r2/r1 = e-aδ
k = 0.060/0.101 = e-0.22*π/2
k = 0.6 =e-0.35
k = 0.6
Since k is our scaling factor then our next edges r3 and r4 as a ratio should be similar
r3 = 0.075e0.22 (1.35-1.57)
r3 = 0.075e-0.05
r3 = 0.07m
Or 70mm